s involving definite integrals (algebraic) AP.CALC: CHA‑4 (EU), CHA‑4.D (LO), CHA‑4.D.1 (EK), CHA‑4.D.2 (EK), CHA‑4.E (LO), CHA‑4.E.1 (EK) Google Classroom Facebook Twitter Email You might need: Calculator Problem The number of hours of daylight in a town changes at a rate of $\dfrac{1}{20}\cos\left(\dfrac{t}{60}+2\right)$ hours per day (where $t$ is the number of days since January $21^{\text{st}}$ ). By approximately how many hours does the daylight increase between $t=240$ and $t=300$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{20}\cos(7)-\dfrac{1}{20}\cos(6)$ (Choice B) B $\dfrac{1}{20}\sin(7)-\dfrac{1}{20}\sin(6)$ (Choice C) C $3\cos(7)-3\cos(6)$ (Choice D) D $3\sin(7)-3\sin(6)$
Letting $h(t)$ be the number of hours of daylight on day $t$, we are given that $h'(t)=\dfrac{1}{20}\cos\left(\dfrac{t}{60}+2\right)$. We want to find $h(300)-h(240)$. According to the Fundamental Theorem of Calculus, $\begin{aligned} h(300)-h(240)&=\int_{240}^{300} h'\left(t\right)dt \\\\ &=\int_{240}^{300}\dfrac{1}{20}\cos\left(\dfrac{t}{60}+2\right)dt \end{aligned}$ $\int_{240}^{300}\dfrac{1}{20}\cos\left(\dfrac{t}{60}+2\right)dt=3\sin(7)-3\sin(6)$ In conclusion, between $t=240$ and $t=300$ the number of daylight hours increases by $3\sin(7)-3\sin(6)$ hours.